Question

How do you test for convergence for `1/((2n+1)!)`?

Answer 1

In the case you meant "test the convergence of the series : `sum_(n=1)^(oo)1/((2n+1)!)`"

the Answer is : it `color(blue)"converges"`

Explanation:

To find out, we can use the ratio test.
That is, if `"U"_"n"` is the `n^"th"` term of this series

Then if, we show that `lim_(nrarr+oo)abs("U"_("n"+1)/"U"_n)<1`
it means that the series converges

On the other if `lim_(nrarr+oo)abs(("U"_("n"+1))/"U"_n)>1`
it means that the series diverges

In our case

`"U"_n=1/((2n+1)!)`

`" "` and

`"U"_("n"+1)=1/([2(n+1)+1]!)=1/([2n+3]!)`

Hence, `"U"_("n"+1)/"U"_n=1/((2n+3)!)÷1/((2n+1)!)=((2n+1)!)/((2n+3)!)`

`"Notice that" :`
`(2n+3)! =(2n+3)xx(2n+2)xx(2n+1)!`

Just like : `10! =10xx9xx8!`
We subtract `1` each time to get the next

So we have,
`"U"_("n"+1)/"U"_n=((2n+1)!)/((2n+3)(2n+2)(2n+1)!)=1/((2n+3)(2n+2))`

Next we test,

`lim_(nrarr+oo)abs("U"_("n"+1)/"U"_n)`

`=lim_(nrarr+oo)abs(1/((2n+3)(2n+2)))=lim_(nrarr+oo)1/((4n^2+10n+6))=1/(+oo)=0" "` and `0` is less than `1`

Hence, it's quite safe to conclude that the series `color(blue)"converges" !`